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up vote 0 down vote favorite 
<?php
include_once 'forecastVo.php';
include_once 'BaseVo.php';
//ini_set("display_errors", 0);
    // The value of the variable name is found
    $count=0;


     $json_url = file_get_contents('http://maps.google.com/maps/api/geocode/json?address='jaipur'&sensor=false'); //line 9
     if($json_url){
    $obj = json_decode($json_url,true);

     $obj2= $obj['results'];

I am developing an application where i need to get the latitude and longitude for a location.

Uptil now this code was working fine..

but now i am getting the error:

Parse error: syntax error, unexpected T_STRING in /home/a4101275/public_html/index.php on line 9

line 9 is where i am using file_get_contents

i tried googling about the same error but could not find a solution? what might be wrong with this?

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You have to use your escape characters correctly. –  Matt Aug 2 '12 at 20:12

3 Answers 3

up vote 3 down vote accepted

You have to use your escape characters correctly. You can't have a single-quote (') inside of a single-quote-encapsulated string. It breaks it. In order to continue the string and have PHP interpret your inner single-quote literally, you have to escape it with \.

$json_url = file_get_contents('http://maps.google.com/maps/api/geocode/json?address=\'jaipur\'&sensor=false');

Or you can use the alternative string encapsulator, double-quote (").

$json_url = file_get_contents("http://maps.google.com/maps/api/geocode/json?address='jaipur'&sensor=false");

For future reference, Parse error: syntax error, unexpected T_STRING usually means you have a bad string somewhere on that line.

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up vote 1 down vote

Why quote it at all? I can't imagine the Google API requires (or even expects) that value to be quoted.

$json_url = file_get_contents('http://maps.google.com/maps/api/geocode/json?address=jaipur&sensor=false'); //line 9

Or, is jaipur a variable? If so:

$json_url = file_get_contents("http://maps.google.com/maps/api/geocode/json?address=$jaipur&sensor=false"); //line 9

Hard to tell from your question what it is you're trying to accomplish...

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up vote 0 down vote 
$json_url = file_get_contents("http://maps.google.com/maps/api/geocode/json?address='jaipur'&sensor=false");

Or escaping it with \

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thanks it works now –  shiven Aug 2 '12 at 20:18

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